Let S={1,2,3,4,.....19,20} Suppose that n is the smallest positive integer

Note that if $ab$ is not a factor of $n$ for relatively prime $a,b>1$ , then at least one of $a,b$ is not a factor of $n$ ; however, since neither of $a,b>1$ is consecutive with $ab$ , this would contradict the definition of $n$ . That is, if $ab \in S$ where $a,b>1$ are relatively prime, then $ab$ must be a factor of $n$ , or in other words, if $m\in S$ is not a factor of $n$ , then $m = p^k$ is the positive power of some prime.

We can even go one step further, as if $p^a \in S$ for some prime $p>1$ and $0\leq b < a$ such that $p^b\in S$ is not a factor of $n$ , then $p^a$ is a multiple of $p^b$ , so $p^a$ is also not a factor of $n$ . As once again we can see that $p^b, p^a$ are not consecutive, this would contradict the definition of $n$ . It follows that if $p^a$ is not a factor of $n$ , then $p^{a+1} > 20$ is not an element of $S$ , so the only possible nonfactors of $n$ are the maximal prime powers of $S$ .

We can list these: $2^4 = 16, 3^2 = 9, 5, 7, 11, 13, 17, 19$ , or in order: $5, 7, 9, 11, 13, 16, 17, 19$ of which the only two consecutive numbers are $16, 17$ so these must be the two nonfactors of $n$ from $S$ . It follows that $n$ is the smallest positive multiple of the elements of $S\setminus \{16,17\}$ , so $n = \text{lcm}(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,18,19,20) = 2^33^25^17^111^113^119^1 = 6846840$ and the sum of the digits is $6+8+4+6+8+4+0 = \boxed{36}$